The document discusses a genetic cross between two heterozygous parents, one exhibiting muppetrus bristle and the other flonasean body color. It concludes that the mode of inheritance for the observed traits is sex-linked recessive, as evidenced by the phenotypic ratios in the offspring. The analysis reinforces that males receive only one X chromosome from the female, resulting in no occurrence of the disease phenotype in females.

1. Directions: Problem 1. A female with Muppetrus bristle mates with a male with Flonasean body
color, producing the following counts: F1 Generation Males Disease-Disease WIT-Disease WT-
WT Disease-WT 243 239 Females WT-Disease Disease Disease WT-WT Disease-WT 254 254
Solution
Parents are heterozygous. Given that each parent has one disease phenotype at one locus and
wild type at other locus.
Female's disease phenotype is Muppetrus bristle. Let its symbol be WT.
Cross is: +/disease (phenotype Muppetrus) crossed to +/disease (phenotype Flonasean)
The answer cannot be homozygous lethal because WT-WT and Disease-Disease are there.
Similarly it cannot be autosomal recessive, because in females Disease-Disease is 0.
This means that the mode of inheritance is sex-linked recessive.
In Drosophila males, crossing over does not occur. Only one X chromosome, which comes from
the female parent. So, genotype of males will be "WT (wild type)" phenotype on X-
chromosome. In females Disease-Disease is zero because of this.